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Principles of Inheritance and Variation: Class 12 Biology Practice Questions

25 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A genetics student, Priya, is studying flower colour inheritance in a plant species. She crosses a true-breeding red-flowered plant with a true-breeding white-flowered plant. All F1 plants show pink flowers. When Priya self-pollinates the F1 plants, she obtains 240 plants in the F2 generation — 60 red-flowered, 120 pink-flowered, and 60 white-flowered.

In a separate experiment, Priya's classmate Rohan studies ABO blood group inheritance in humans. He notes that his father has blood group A (genotype I^A i) and his mother has blood group B (genotype I^B i). Rohan wants to predict all possible blood groups in their children.

Read the following passage carefully and answer the questions that follow:

A genetics student, Priya, is studying flower colour inheritance in a plant species. She crosses a true-breeding red-flowered plant with a true-breeding white-flowered plant. All F1 plants show pink flowers. When Priya self-pollinates the F1 plants, she obtains 240 plants in the F2 generation — 60 red-flowered, 120 pink-flowered, and 60 white-flowered.

In a separate experiment, Priya's classmate Rohan studies ABO blood group inheritance in humans. He notes that his father has blood group A (genotype I^A i) and his mother has blood group B (genotype I^B i). Rohan wants to predict all possible blood groups in their children.

(a) Identify the type of inheritance observed in Priya's experiment. Give ONE reason to justify your answer. (1 mark)
(b) Write the genotypes of the true-breeding red-flowered and white-flowered parent plants. Using a Punnett square, show the F1 and F2 generations, and state the genotypic and phenotypic ratios obtained in F2. (2 marks)
(c) From Rohan's family cross (I^A i × I^B i), list ALL possible blood groups (phenotypes) that can appear in their children. Which blood group demonstrates co-dominance, and why? (1 mark)

Show answer
MARKING SCHEME (Total: 4 marks)

─────────────────────────────────────────
(a) [1 mark]
─────────────────────────────────────────
• Type of inheritance: Incomplete dominance [½ mark]
• Justification: The F1 heterozygote (Rr) shows an INTERMEDIATE phenotype (pink) instead of expressing either parental phenotype fully; neither allele completely masks the other. [½ mark]

─────────────────────────────────────────
(b) [2 marks]
─────────────────────────────────────────
Genotypes of parents:
• True-breeding red-flowered parent: RR
• True-breeding white-flowered parent: WW
(Using R for red allele and W/r for white allele — standard NCERT notation: R¹R¹ × R²R² or RR × rr)

For clarity, using: R = red allele, r = white allele
Parent (P): RR (red) × rr (white) [½ mark]

F1 Generation:
All F1 = Rr (PINK) — due to incomplete dominance

F1 self-pollination cross: Rr × Rr

GAMETES OF F1 PARENTS:
Rr parent produces gametes: R and r

PUNNETT SQUARE (F2): [½ mark]

┌─────────────┬──────────────┬──────────────┐
│ Rr × Rr │ R (gamete) │ r (gamete) │
├─────────────┼──────────────┼──────────────┤
│ R (gamete) │ RR │ Rr │
├─────────────┼──────────────┼──────────────┤
│ r (gamete) │ Rr │ rr │
└─────────────┴──────────────┴──────────────┘

F2 Results:
RR = Red-flowered → 1
Rr = Pink-flowered → 2
rr = White-flowered → 1

Genotypic ratio (F2): 1 RR : 2 Rr : 1 rr [¼ mark]
Phenotypic ratio (F2): 1 Red : 2 Pink : 1 White [¼ mark]

Verification with Priya's data:
Total = 240 plants → 60 RR (red) : 120 Rr (pink) : 60 rr (white) = 1:2:1 ✓

Note: In incomplete dominance, phenotypic ratio = genotypic ratio (BOTH are 1:2:1), unlike Mendelian dominance where F2 phenotypic ratio is 3:1.

─────────────────────────────────────────
(c) [1 mark]
─────────────────────────────────────────
Cross: Father I^A i × Mother I^B i

Gametes:
Father produces: I^A and i
Mother produces: I^B and i

Possible genotypes in children:
I^A I^B → Blood group AB
I^A i → Blood group A
I^B i → Blood group B
ii → Blood group O

All four possible blood groups in their children: A, B, AB, O [½ mark]

Blood group demonstrating co-dominance: AB (genotype I^A I^B) [½ mark]
Reason: In the I^A I^B individual, BOTH alleles I^A and I^B are expressed simultaneously and equally — I^A directs synthesis of antigen A and I^B directs synthesis of antigen B on the RBC surface. Neither allele is dominant over the other; both are fully expressed in the heterozygote — this is co-dominance.

─────────────────────────────────────────
KEY DISTINCTIONS (for examiner reference):
• Incomplete dominance: F1 shows INTERMEDIATE phenotype; F2 ratio = 1:2:1 (both phenotypic and genotypic).
• Co-dominance: BOTH parental phenotypes expressed simultaneously in F1 heterozygote (e.g., AB blood group shows BOTH A and B antigens).
• Multiple alleles: ABO blood group system has THREE alleles (I^A, I^B, i) for a single gene locus — only two present in any one individual.
─────────────────────────────────────────
Q2Case-based4 marks

A genetics teacher presented the following family pedigree to her Class XII students and asked them to analyse it:

Generation I: Carrier female (I-1) × Unaffected male (I-2)
Generation II: Affected male (II-1), Unaffected female (II-2), Unaffected male (II-3)
Generation III: Affected male (III-1) [son of II-2 and her unaffected husband]

The trait shown in the pedigree skips generations, affects more males than females, and unaffected females can pass it to their sons. The teacher asked the students to answer the following questions based on the pedigree.

A genetics teacher presented the following family pedigree to her Class XII students and asked them to analyse it:

Generation I: Carrier female (I-1) × Unaffected male (I-2)
Generation II: Affected male (II-1), Unaffected female (II-2), Unaffected male (II-3)
Generation III: Affected male (III-1) [son of II-2 and her unaffected husband]

The trait shown in the pedigree skips generations, affects more males than females, and unaffected females can pass it to their sons.

(a) Identify the pattern of inheritance shown in this pedigree. Give TWO reasons to justify your answer. (2 marks)
(b) Write the genotypes of Individual I-1 (carrier female) and Individual II-1 (affected male) using proper genetic notation. (1 mark)
(c) Individual II-2 (unaffected female from Generation II) marries an unaffected male. Draw a Punnett square to show the possible genotypes of their children and state the probability that their son will be affected. (1 mark)

Diagram for question 2: Principles of Inheritance and Variation
Show answer
CBSE MARKING SCHEME — 4 Marks

━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Part (a): Pattern of inheritance + Two justifications [2 marks]
━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Pattern of inheritance: X-linked recessive inheritance. (1 mark)

Justification (any TWO of the following, ½ mark each — but awarded as 1 mark total for two valid reasons):

Reason 1: The trait skips Generation I females (I-1 is unaffected but a carrier), indicating the allele is recessive and carried on the X chromosome — if it were autosomal recessive, the carrier status would not be sex-linked. (½)

Reason 2: More males are affected than females (II-1 and III-1 are both males) — males have only one X chromosome (hemizygous), so a single recessive allele on the X chromosome is sufficient to express the trait, whereas females need two copies. (½)

Reason 3: The trait is transmitted from unaffected carrier mother (I-1) to affected son (II-1) — this criss-cross inheritance pattern is characteristic of X-linked recessive traits. (½)

Reason 4: Unaffected females (II-2) in Generation II can pass the trait to their sons (III-1), confirming maternal transmission via the X chromosome. (½)

[Award 1 mark for any TWO valid reasons; 1 mark for correctly identifying X-linked recessive]
Total for (a): 2 marks

━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Part (b): Genotypes of I-1 and II-1 [1 mark]
━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Let X^H = dominant (normal) allele; X^h = recessive (disease) allele

• Individual I-1 (carrier female): X^H X^h
• Individual II-1 (affected male): X^h Y

(Award 1 mark only if BOTH genotypes are correctly written with proper X-linked notation)
Total for (b): 1 mark

━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Part (c): Punnett square + probability [1 mark]
━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Since II-1 (affected male) is in Generation II and II-2 is unaffected, II-2 must be a carrier (X^H X^h) — she has an affected brother (II-1) and an affected son (III-1).

Cross: Carrier female (II-2) × Unaffected male (her husband)
X^H X^h × X^H Y

Parental gametes:
• Female gametes: X^H , X^h
• Male gametes: X^H , Y

Punnett Square:

┌─────────────┬─────────────┬─────────────┐
│ │ X^H │ Y │
│ (father) │ (gamete) │ (gamete) │
├─────────────┼─────────────┼─────────────┤
│ X^H (mother)│ X^H X^H │ X^H Y │
│ │(unaffected │(unaffected │
│ │ female) │ male) │
├─────────────┼─────────────┼─────────────┤
│ X^h (mother)│ X^H X^h │ X^h Y │
│ │ (carrier │ (affected │
│ │ female) │ male) │
└─────────────┴─────────────┴─────────────┘

Offspring genotypes:
1. X^H X^H — Normal female (25%)
2. X^H X^h — Carrier female (25%)
3. X^H Y — Normal male (25%)
4. X^h Y — Affected male (25%)

Probability that their SON will be affected:
Out of all sons (X^H Y and X^h Y), half are affected.
∴ Probability = 1/2 (50%)

(Award 1 mark for correct Punnett square with genotypes AND correct probability stated)
Total for (c): 1 mark

━━━━━━━━━━━━━━━━━━━━━━━━━━━━
TOTAL: 4 marks
━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Q3Case-based4 marks

A genetics counsellor is examining the family history of the Sharma family. Mrs. Sharma has normal vision, but her father was colour blind. Mr. Sharma has normal vision and no history of colour blindness in his family. They have three children: a colour-blind son (Arjun), a daughter with normal vision (Priya), and another son with normal vision (Rohit).

Colour blindness is an X-linked recessive trait. The allele for normal vision is represented as X^N and the allele for colour blindness is represented as X^n.

Read the following case and answer the questions that follow:

A genetics counsellor is examining the family history of the Sharma family. Mrs. Sharma has normal vision, but her father was colour blind. Mr. Sharma has normal vision and no history of colour blindness in his family. They have three children: a colour-blind son (Arjun), a daughter with normal vision (Priya), and another son with normal vision (Rohit).

Colour blindness is an X-linked recessive trait. The allele for normal vision is represented as X^N and the allele for colour blindness is represented as X^n.

(a) What is the genotype of Mrs. Sharma? Give a reason for your answer. [1]
(b) Show with the help of a Punnett square the cross between Mr. and Mrs. Sharma. [2]
(c) What is the probability that Priya is a carrier of colour blindness? Justify your answer. [1]

Show answer
MARKING SCHEME

(a) Genotype of Mrs. Sharma: X^N X^n (Carrier / heterozygous female) [1 mark]

Reason: Mrs. Sharma has normal vision (so she must have at least one X^N allele). Her father was colour blind, meaning his genotype was X^n Y. He could only pass his X^n allele to his daughter (Mrs. Sharma). Therefore, Mrs. Sharma must have received X^n from her father and X^N from her mother, making her genotype X^N X^n (carrier).

──────────────────────────────────────────

(b) Punnett Square for the cross: [2 marks]

Parents:
Mother (Mrs. Sharma) — Carrier female: X^N X^n
Father (Mr. Sharma) — Normal male: X^N Y

Gametes:
Mrs. Sharma produces: X^N and X^n
Mr. Sharma produces: X^N and Y

Punnett Square:

┌───────────────┬───────────────┬───────────────┐
│ │ X^N (Mrs.) │ X^n (Mrs.) │
├───────────────┼───────────────┼───────────────┤
│ X^N (Mr.) │ X^N X^N │ X^N X^n │
│ │ (Normal ♀) │ (Carrier ♀) │
├───────────────┼───────────────┼───────────────┤
│ Y (Mr.) │ X^N Y │ X^n Y │
│ │ (Normal ♂) │(Colour-blind ♂)│
└───────────────┴───────────────┴───────────────┘

Offspring genotypes and phenotypes:
• X^N X^N — Normal vision female (25%)
• X^N X^n — Carrier female, normal vision (25%)
• X^N Y — Normal vision male (25%)
• X^n Y — Colour-blind male (25%)

Phenotypic ratio:
Females: 1 normal : 1 carrier (both appear normal)
Males: 1 normal : 1 colour-blind

[Award 1 mark for correct gametes and parental genotypes written above the grid; 1 mark for the correctly filled Punnett square with genotypes/phenotypes]

──────────────────────────────────────────

(c) Probability that Priya is a carrier: 1/2 (50%) [1 mark]

Justification: From the Punnett square, daughters of this cross can be either X^N X^N (normal, non-carrier) or X^N X^n (carrier) in equal proportion (1 : 1). Since Priya has normal vision (ruling out X^n X^n, which is impossible here anyway), she is either X^N X^N or X^N X^n. Therefore, the probability that Priya is a carrier (X^N X^n) is 1/2 or 50%.
Q4Case-based4 marks

A genetics teacher showed her students a pedigree chart of a family across three generations. In Generation I, a normal male married a normal female. In Generation II, they had two sons and two daughters — one son was colour blind, the other son was normal, one daughter was normal (who later married a normal male), and the other daughter was normal. In Generation III, the normal daughter from Generation II had two children — a colour blind son and a normal daughter.

The teacher then asked the students to analyse the pattern of inheritance and draw the relevant genetic cross.

Read the following case and answer the questions that follow:

A genetics teacher showed her students a pedigree chart of a family across three generations. In Generation I, a normal male married a normal female. In Generation II, they had two sons and two daughters — one son was colour blind, the other son was normal, one daughter was normal (who later married a normal male), and the other daughter was normal. In Generation III, the normal daughter from Generation II had two children — a colour blind son and a normal daughter.

The teacher then asked the students to analyse the pattern of inheritance and draw the relevant genetic cross.

(i) Identify the pattern of inheritance shown in the pedigree chart. Give one reason to support your answer. (1 mark)

(ii) What would be the genotype of the Generation I female (mother)? Justify your answer. (1 mark)

(iii) Draw a labelled Punnett square showing the cross between the Generation II normal daughter and her normal husband. Show the genotypes of all offspring. (2 marks)

Show answer
MARKING SCHEME — CASE STUDY (Total: 4 marks)

─────────────────────────────────────────
(i) Pattern of inheritance: (1 mark)
─────────────────────────────────────────
Value point: X-linked recessive inheritance.

Reason (any one acceptable):
• Only males are affected (colour blind son in Gen II and colour blind grandson in Gen III); females are unaffected carriers — characteristic of X-linked recessive traits.
• The trait skips a generation in females (Generation I female is normal but passes the trait to her son and to her daughter who passes it further) — typical of X-linked recessive pattern.

[1 mark for correct identification with valid reason]

─────────────────────────────────────────
(ii) Genotype of Generation I female: (1 mark)
─────────────────────────────────────────
Notation used: X^C = normal allele (dominant); X^c = colour blind allele (recessive)

Genotype of Generation I female = X^C X^c (Carrier female)

Justification: She is phenotypically normal (not colour blind), so she must have at least one X^C allele. However, she has a colour blind son (X^c Y), who must have received X^c from his mother (since father contributes Y to sons). Therefore, the mother must be a carrier: X^C X^c.

[1 mark for correct genotype with valid justification]

─────────────────────────────────────────
(iii) Punnett Square — Cross between Generation II normal daughter (carrier) × normal male: (2 marks)
─────────────────────────────────────────

Parents:
Mother (Gen II normal daughter) = X^C X^c (carrier female)
Father (normal male) = X^C Y

Gametes:
Mother produces: X^C and X^c
Father produces: X^C and Y

Punnett Square:

┌──────────────┬──────────────┬──────────────┐
│ │ X^C │ Y │
│ (Father →) │ (gamete 1) │ (gamete 2) │
├──────────────┼──────────────┼──────────────┤
│ X^C │ X^C X^C │ X^C Y │
│ (gamete 1) │ Normal female│ Normal male │
├──────────────┼──────────────┼──────────────┤
│ X^c │ X^C X^c │ X^c Y │
│ (gamete 2) │Carrier female│Colour blind │
│ │ (normal) │ male │
└──────────────┴──────────────┴──────────────┘

Genotypes and Phenotypes of offspring:
• X^C X^C — Normal female (non-carrier) — 25%
• X^C X^c — Carrier female (phenotypically normal) — 25%
• X^C Y — Normal male — 25%
• X^c Y — Colour blind male — 25%

Phenotypic ratio: 3 normal : 1 colour blind (among all offspring)
50% daughters normal : 50% sons normal, 50% sons colour blind

[1 mark for correct Punnett square with parental gametes on top row and left column]
[1 mark for correct genotypes/phenotypes of all four offspring labelled]

Total: 4 marks (1 + 1 + 2)
Q5MCQ1 mark

Which of the following statements about linkage and recombination is INCORRECT?

Show answer
Correct answer: (B)

Recombination frequency between two genes on the same chromosome CANNOT exceed 50%. When two genes are very far apart on the same chromosome, multiple crossovers occur between them, and the recombination frequency approaches — but never exceeds — 50%. A value of 50% recombination is indistinguishable from independent assortment (as if the genes were on separate chromosomes). Therefore, the statement that recombination frequency 'can exceed 50%' is incorrect.

• Option (A) is CORRECT: Genes on the same chromosome are physically linked and tend to be inherited together (linked genes — Morgan's work with Drosophila).
• Option (C) is CORRECT: Morgan's experiments with Drosophila showed that X-linked genes such as those for body colour and wing size exhibited linkage — they were inherited together more frequently than expected under Mendel's law of independent assortment.
• Option (D) is CORRECT: 1 centiMorgan (cM) = 1 map unit = 1% recombination frequency between two loci; this is the standard unit of genetic map distance.
Q6Short Answer1 mark

Assertion (A): In a cross between a carrier female (X^H X^h) and a normal male (X^H Y), 50% of the sons are expected to be haemophilic.
Reason (R): Haemophilia is an X-linked recessive disorder in which the defective allele is transmitted from carrier mother to sons through criss-cross inheritance.

Show answer
Correct option: (A) Both A and R are true, and R is the correct explanation of A.

Explanation:

Assertion is TRUE:
Cross: Carrier female (X^H X^h) × Normal male (X^H Y)

Gametes of female: X^H , X^h
Gametes of male: X^H , Y

Punnett Square:

| | X^H (from father) | Y (from father) |
|----------|--------------------|------------------|
| X^H (mother) | X^H X^H (normal female) | X^H Y (normal male) |
| X^h (mother) | X^H X^h (carrier female) | X^h Y (haemophilic male) |

Sons produced: X^H Y (normal) : X^h Y (haemophilic) = 1 : 1
→ 50% of sons are haemophilic. ✓

Reason is TRUE and correctly explains A:
Haemophilia is caused by a recessive allele (X^h) located on the X chromosome. Males are hemizygous (only one X chromosome), so a single copy of X^h causes the disease. The carrier mother (X^H X^h) passes the X^h allele to 50% of her sons — this pattern of transmission from mother to son is called criss-cross inheritance. This is precisely why 50% of sons are affected. ✓

∴ R is the correct and complete explanation of A. → Option (A) is correct.
Q7MCQ1 mark

A farmer notices that when he crosses two true-breeding varieties of snapdragon (Antirrhinum majus) — one with red flowers and one with white flowers — all F1 plants have pink flowers. He then self-pollinates the F1 plants to obtain F2. A pest destroys all white-flowered F2 plants before he can count them. Which of the following correctly identifies the genotypic ratio of the SURVIVING F2 plants?

Show answer
Correct answer: (B) 1 RR : 2 Rr

Reasoning (for examiner reference):

• Snapdragon flower colour shows INCOMPLETE DOMINANCE — F1 heterozygote (Rr) is pink, not red. RR = red, Rr = pink, rr = white.

• Cross: P → RR (red) × rr (white)
F1 → all Rr (pink)

• F1 self: Rr × Rr

Punnett square (F2):

| | R | r |
|--------|--------|--------|
| R | RR | Rr |
| r | Rr | rr |

F2 genotypic ratio: 1 RR : 2 Rr : 1 rr
F2 phenotypic ratio: 1 red : 2 pink : 1 white

• The pest destroys all rr (white) plants. Surviving plants = RR + Rr only.

• Genotypic ratio of survivors = 1 RR : 2 Rr

• Why the other options are wrong:
(A) 1 RR : 1 Rr — would only occur if all heterozygotes had also been selectively removed, leaving equal numbers; incorrect.
(C) 2 RR : 1 Rr — inverts the correct ratio; incorrect.
(D) 1 RR : 1 Rr : 1 rr — includes rr plants which were destroyed; incorrect.
Q8Short Answer1 mark

Assertion (A): In humans, the sex of the offspring is determined by the type of sperm that fertilises the egg.
Reason (R): Females are homogametic (XX) and produce only one type of egg, while males are heterogametic (XY) and produce two types of sperm.

Show answer
Correct option: (A) Both A and R are true, and R is the correct explanation of A.

Explanation:
• Assertion is TRUE: In humans (XX/XY sex determination system), the sperm — not the egg — determines the sex of the offspring, because the egg always contributes one X chromosome, whereas the sperm may contribute either X or Y.
• Reason is TRUE: Human females are homogametic — they carry two X chromosomes (XX) and therefore produce only one type of egg (22 + X). Human males are heterogametic — they carry one X and one Y chromosome (XY) and therefore produce two types of sperm: 22 + X (gives female offspring) and 22 + Y (gives male offspring).
• R correctly explains A: Because only the male produces two genetically different types of gametes, it is the type of sperm (X-bearing or Y-bearing) that fertilises the egg which determines whether the offspring will be female (XX) or male (XY). Thus R is the direct mechanistic explanation of A.

[1 mark]
Q9MCQ1 mark

A researcher studying a rare autosomal recessive disorder in a family notices the following: the affected child (III-1) has two phenotypically normal parents (II-1 and II-2). The maternal grandfather (I-1) is affected, but the maternal grandmother (I-2) and paternal grandparents (I-3 and I-4) are all phenotypically normal. The researcher correctly predicts the probability that the next child born to II-1 and II-2 will be an affected daughter.

Which of the following gives the correct genotypes of II-1 and II-2, AND the probability of an affected daughter in their next child?

Show answer
Correct Answer: (B) II-1 = Aa, II-2 = Aa; probability of affected daughter = 1/8

Reasoning (examiner's value points):

1. Mode of inheritance:
• Disorder is autosomal recessive (let 'a' = recessive disease allele; 'A' = dominant normal allele).
• Autosomal: trait appears in both sexes; affected individuals are homozygous recessive (aa).

2. Genotype of II-1 (mother of affected child):
• Maternal grandfather I-1 is affected → genotype = aa.
• I-1 must have passed one 'a' allele to his daughter II-1.
• II-1 is phenotypically normal → must carry at least one 'A'.
• Therefore II-1 = Aa (obligate carrier).

3. Genotype of II-2 (father of affected child):
• II-2 is phenotypically normal but has an affected child (aa).
• To contribute 'a' to the affected child, II-2 must carry 'a'.
• Therefore II-2 = Aa (obligate carrier).
• (Note: paternal grandparents I-3 and I-4 are normal; II-2 could have received 'a' from either carrier grandparent.)

4. Cross between II-1 (Aa) × II-2 (Aa):

Punnett Square:

| | A (II-2) | a (II-2) |
|---------|----------|----------|
| A (II-1)| AA | Aa |
| a (II-1)| Aa | aa |

Genotypic ratio: 1 AA : 2 Aa : 1 aa
Phenotypic ratio: 3 normal : 1 affected
Probability of affected child (aa) = 1/4

5. Probability of an affected DAUGHTER specifically:
• Disorder is autosomal → sex of child is independent of genotype.
• P(affected, i.e., aa) = 1/4
• P(daughter) = 1/2
• P(affected daughter) = 1/4 × 1/2 = 1/8

Why other options are wrong:
(A) Gives correct genotypes but wrong probability (1/4 gives probability of any affected child, not specifically a daughter).
(C) II-1 cannot be aa — she is phenotypically normal.
(D) If II-2 = AA, they cannot produce an affected child (aa), which contradicts the given data (III-1 is affected).
Q10MCQ1 mark

A farmer crosses two true-breeding pea plants — one with round seeds (RR) and one with wrinkled seeds (rr). He plants all the F1 seeds and allows them to self-pollinate to get F2 seeds. He then randomly picks ONE seed from the F2 generation. What is the probability that this seed is wrinkled?

Show answer
Correct answer: (B) 1/4

Reasoning (step-by-step):

Step 1 — Parental Cross (P generation):
RR (round) × rr (wrinkled)
Gametes: R (from RR) and r (from rr)
F1 genotype: All Rr (round — R is dominant over r)

Step 2 — F1 Self-pollination (F1 × F1):
Rr × Rr

Step 3 — F2 Punnett Square:

| | R | r |
|---------|---------|---------|
| R | RR | Rr |
| r | Rr | rr |

F2 genotypic ratio: 1 RR : 2 Rr : 1 rr
F2 phenotypic ratio: 3 Round (RR + Rr) : 1 Wrinkled (rr)

Step 4 — Probability of wrinkled seed (rr) in F2:
= 1 out of 4 = 1/4

Key concept applied: Law of Segregation (Mendel's First Law) — alleles separate during gamete formation; R (round) is dominant over r (wrinkled); wrinkled phenotype appears only in homozygous recessive (rr) condition.

Hence, the correct answer is (B) 1/4.
Q11Short Answer2 marks

Study the incomplete Punnett square given below showing a monohybrid cross between two pea plants heterozygous for seed colour (Yellow, Y is dominant over green, y).

Y y
┌─────────┬─────────┐
Y │ YY │ │
├─────────┼─────────┤
y │ │ yy │
└─────────┴─────────┘

(a) Complete the Punnett square by filling in the two missing genotypes.
(b) What will be the phenotypic ratio of the offspring from this cross?

Diagram for question 11: Principles of Inheritance and Variation
Show answer
(a) Completed Punnett square:

Parents: Yy × Yy
Gametes: Y, y × Y, y

Y y
┌──────────┬──────────┐
Y │ YY │ Yy │
├──────────┼──────────┤
y │ Yy │ yy │
└──────────┴──────────┘

Missing genotypes filled in:
• Top-right cell: Yy
• Bottom-left cell: Yy
(1 mark — both missing genotypes correctly filled)

(b) Phenotypic ratio:
• YY and Yy → Yellow seeds (3 plants)
• yy → green seeds (1 plant)
• Phenotypic ratio = 3 Yellow : 1 green
(1 mark)

(1 × 2 = 2 marks)
Q12Short Answer2 marks

A genetics student performed two separate monohybrid crosses using pea plants:

Cross P: Tall (TT) × Dwarf (tt) → All F1 plants were Tall
Cross Q: When two F1 plants from Cross P were crossed with each other, the F2 generation showed both Tall and Dwarf plants.

(a) What principle of Mendel does the result of Cross P demonstrate? (1 mark)
(b) In Cross Q, a student claims that the ratio of Tall : Dwarf plants in F2 will be 3:1. Is the student correct? Justify your answer in one sentence. (1 mark)

Show answer
(a) The result of Cross P demonstrates Mendel's Law of Dominance — when parents differing in a pair of contrasting characters are crossed, the character that appears in the F1 generation (Tall) is the dominant character, and the one that is masked (Dwarf) is the recessive character. (1 mark)

(b) Yes, the student is correct. When two F1 heterozygotes (Tt × Tt) are crossed, the F2 genotypic ratio is 1 TT : 2 Tt : 1 tt, giving a phenotypic ratio of 3 Tall : 1 Dwarf, because the 'T' allele is dominant over 't'. (1 mark)

[Examiner note: Award 1 mark for correct principle name in (a). Award 1 mark in (b) only if the student states BOTH that the claim is correct AND provides a valid genetic justification referencing heterozygous F1 (Tt) or the Punnett square logic. A bare 'yes' without justification = 0 marks for (b).]
Q13Short Answer2 marks

A woman with normal vision but whose father was colour blind marries a man with normal vision. What is the probability that their son will be colour blind? Give the genotypes of both parents and justify your answer.

Show answer
Genotype of mother (carrier female): X^C X^c [½ mark]
(Her father was colour blind, so he had genotype X^c Y; she received X^c from father → she is a carrier)

Genotype of father (normal vision male): X^C Y [½ mark]

Cross:

| | X^C (from father) | Y (from father) |
|---|---|---|
| X^C (from mother) | X^C X^C (normal female) | X^C Y (normal male) |
| X^c (from mother) | X^C X^c (carrier female) | X^c Y (colour blind male) |

Probability that their SON will be colour blind = 1/2 (50%) [½ mark]

Justification: Sons receive the Y chromosome from father and either X^C or X^c from the carrier mother. A son who receives X^c from mother will be colour blind (X^c Y), since colour blindness is an X-linked recessive trait and males are hemizygous — a single recessive allele on the X chromosome is sufficient to express the trait. [½ mark]
Q14Short Answer2 marks

A genetics student is analysing the pedigree of a family in which a male child is born with Klinefelter syndrome. During meiosis in the mother, a non-disjunction event occurred.
(i) Write the genotype (in chromosomal notation) of this affected child.
(ii) Identify the specific stage of maternal meiosis at which non-disjunction must have occurred to produce this child, and justify your answer in one sentence.

Show answer
(i) Genotype of the affected child:
44 + XXY (i.e., 2n = 47; 44 autosomes + two X chromosomes + one Y chromosome) (1)

(ii) Stage of maternal meiosis and justification:

Meiosis I — Non-disjunction of the two homologous X chromosomes during Meiosis I produces a secondary oocyte carrying both X chromosomes, which after normal Meiosis II yields an XX-bearing egg; fertilisation of this egg by a Y-bearing sperm gives the 44+XXY Klinefelter karyotype. (1)

OR

Meiosis II — Non-disjunction of the sister chromatids of the X chromosome during Meiosis II produces an XX-bearing egg directly; fertilisation of this egg by a Y-bearing sperm gives the 44+XXY Klinefelter karyotype. (1)
Q15Short Answer2 marks

In humans, a person is found to have 47 chromosomes with the karyotype 45 + XXY. (i) Name the chromosomal disorder this person is suffering from. (ii) State one characteristic feature of this disorder.

Show answer
(i) The chromosomal disorder is Klinefelter's Syndrome. (1 mark)

(ii) Any ONE characteristic feature (1 mark):
• The affected individual is phenotypically male but has small, underdeveloped testes (testicular dysgenesis) and is sterile / infertile.
OR
• The individual shows gynaecomastia (development of breast-like tissue in males).
OR
• The individual has an overall masculine appearance but may show some feminine characteristics and is usually mentally normal or mildly intellectually impaired.

[Note to examiner: Award 1 mark for correct name in (i). Award 1 mark for any one correct feature in (ii). Karyotype = 2n + 1 = 47 (trisomy of sex chromosomes — two X and one Y).]
Q16Short Answer3 marks

A genetics student is analysing a pedigree of a family in which a particular disorder appears. She observes the following: (i) The disorder appears in every generation. (ii) Both males and females are affected with equal frequency. (iii) An affected father has an affected son. Based on these observations, she concludes that the disorder follows autosomal dominant inheritance. However, her friend argues that the disorder could also be X-linked dominant. The student then finds one additional piece of evidence that conclusively rules out X-linked dominant inheritance. (a) Identify the ONE piece of evidence (from the three observations given) that conclusively rules out X-linked dominant inheritance. Give a reason. (b) Work out the cross between an affected heterozygous father (autosomal dominant) and a normal mother to show the expected genotypic and phenotypic ratios in the offspring.

Show answer
PART (a) — 1 mark

The observation that conclusively rules out X-linked dominant inheritance is:
Observation (iii): An affected father has an affected son.

Reason: In X-linked dominant inheritance, a father (X^A Y) passes his X chromosome ONLY to daughters, never to sons (sons receive Y from father). Therefore, an affected father CANNOT transmit an X-linked dominant allele to his son. Since the son IS affected, the gene cannot be on the X chromosome. This rules out X-linked dominant inheritance conclusively. ✓ (1 mark)

─────────────────────────────────────────
PART (b) — 2 marks

Let dominant allele (disorder) = A; recessive allele (normal) = a

Parents:
• Affected heterozygous father × Normal mother
• Genotype: Aa × aa

Gametes produced:
• Father's gametes: A or a
• Mother's gametes: a or a

Punnett Square:

┌─────────────┬─────────────┬─────────────┐
│ Father → │ A │ a │
│ Mother ↓ │ │ │
├─────────────┼─────────────┼─────────────┤
│ a │ Aa │ aa │
├─────────────┼─────────────┼─────────────┤
│ a │ Aa │ aa │
└─────────────┴─────────────┴─────────────┘

Offspring genotypes: Aa : aa = 1 : 1

Genotypic ratio: 1 Aa : 1 aa (i.e., 2 Aa : 2 aa out of 4) ✓ (½ mark)

Phenotypic ratio:
• Aa → Affected (dominant allele present)
• aa → Normal

Phenotypic ratio: 1 Affected : 1 Normal ✓ (½ mark)

∴ 50% offspring are expected to be affected (both males and females equally). ✓ (1 mark for correct phenotypic interpretation)

─────────────────────────────────────────
MARK SUMMARY:
• Part (a) correct identification + valid reason = 1 mark
• Correct genotypes of parents and gametes shown = ½ mark
• Correct Punnett square (all 4 cells) = ½ mark
• Correct genotypic ratio = ½ mark
• Correct phenotypic ratio with interpretation = ½ mark
Total = 3 marks
Q17Short Answer3 marks

Priya is a healthy woman whose brother has sickle cell anaemia. She wants to know whether she is a carrier of the sickle cell trait before planning her family. Her husband Arjun has no family history of the disease.

(a) State the inheritance pattern of sickle cell anaemia and identify Priya's possible genotypes. (1 mark)
(b) If Priya is confirmed to be a carrier and she marries Arjun (who is normal, non-carrier), draw a cross to show the expected genotypes and phenotypes of their offspring. State the genotypic ratio. (2 marks)

Show answer
CBSE MARKING SCHEME — 3 marks total

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
PART (a) — 1 mark
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

• Inheritance pattern: Sickle cell anaemia is an autosomal recessive disorder. [½ mark]
(Gene located on chromosome 11; caused by point mutation GAG → GTG in β-globin gene → Glu → Val substitution)

• Priya's possible genotypes:
— Priya is phenotypically NORMAL but has an affected brother, so she could be either:
(i) Hb^A Hb^A — homozygous normal
(ii) Hb^A Hb^S — carrier (heterozygous) [½ mark]
(She cannot be Hb^S Hb^S as she is healthy.)

[Notation: Hb^A = normal allele; Hb^S = sickle cell allele]

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
PART (b) — 2 marks
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Parents: Priya (Carrier) × Arjun (Normal, non-carrier)
Genotypes: Hb^A Hb^S × Hb^A Hb^A

Gametes: Hb^A or Hb^S Hb^A only

PUNNETT SQUARE:

┌─────────────────┬──────────────────┬──────────────────┐
│ │ Gamete: Hb^A │ Gamete: Hb^A │
│ (from Arjun) │ │ │
├─────────────────┼──────────────────┼──────────────────┤
│ Gamete: Hb^A │ Hb^A Hb^A │ Hb^A Hb^A │
│ (from Priya) │ (Normal) │ (Normal) │
├─────────────────┼──────────────────┼──────────────────┤
│ Gamete: Hb^S │ Hb^A Hb^S │ Hb^A Hb^S │
│ (from Priya) │ (Carrier) │ (Carrier) │
└─────────────────┴──────────────────┴──────────────────┘

[Award 1 mark for correctly drawn Punnett square with gametes on top row and left column]

Genotypic ratio:
Hb^A Hb^A : Hb^A Hb^S = 2 : 2 = 1 : 1 [½ mark]

Phenotypic outcome:
— 2 offspring normal (Hb^A Hb^A)
— 2 offspring carriers but phenotypically normal (Hb^A Hb^S)
— NO child will have sickle cell anaemia (Hb^S Hb^S) [½ mark]

[Value point: Since Arjun is non-carrier, none of the offspring can be affected; maximum risk is 50% chance of being a carrier.]

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
MARK ALLOCATION SUMMARY:
(a) Autosomal recessive pattern — ½ mark
Priya's two possible genotypes — ½ mark
(b) Correct Punnett square — 1 mark
Genotypic ratio stated correctly — ½ mark
Phenotypic conclusion — ½ mark
TOTAL = 3 marks
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Q18Short Answer3 marks

A genetics student performed a dihybrid cross between two true-breeding plants — one with round, yellow seeds (RRYY) and another with wrinkled, green seeds (rryy). The F₁ plants were then test-crossed.

(a) Write the genotype and phenotype of the F₁ generation. (1 mark)
(b) Give the genotypic and phenotypic ratios expected in the test-cross progeny, assuming both genes are on different (non-homologous) chromosomes. (1 mark)
(c) If, instead, the two genes were found to be located on the same chromosome and showed incomplete linkage, how would the test-cross results differ from those obtained in (b)? Give one reason for your answer. (1 mark)

Show answer
CBSE Marking Scheme — 3 Marks

─────────────────────────────────────
(a) Genotype and Phenotype of F₁ (1 mark)
─────────────────────────────────────
• Cross: RRYY (round, yellow) × rryy (wrinkled, green)
• F₁ genotype: RrYy [½ mark]
• F₁ phenotype: Round, Yellow seeds (dominant phenotype expressed for both traits) [½ mark]

─────────────────────────────────────
(b) Test-cross of F₁ (RrYy × rryy) — genes on non-homologous chromosomes (1 mark)
─────────────────────────────────────
Test-cross parent gametes: ry only
F₁ gametes (independent assortment): RY, Ry, rY, ry (each 25%)

Test-cross Punnett Grid:

┌──────────────┬─────────────┐
│ F₁ gametes │ rryy parent│
│ │ (ry) │
├──────────────┼─────────────┤
│ RY │ RrYy │
├──────────────┼─────────────┤
│ Ry │ Rryy │
├──────────────┼─────────────┤
│ rY │ rrYy │
├──────────────┼─────────────┤
│ ry │ rryy │
└──────────────┴─────────────┘

• Genotypic ratio: 1 RrYy : 1 Rryy : 1 rrYy : 1 rryy [½ mark]
• Phenotypic ratio: 1 Round Yellow : 1 Round Green : 1 Wrinkled Yellow : 1 Wrinkled Green
(i.e., 1 : 1 : 1 : 1) [½ mark]

─────────────────────────────────────
(c) Effect of incomplete linkage on test-cross results (1 mark)
─────────────────────────────────────
• If the two genes are on the SAME chromosome (linked) with incomplete linkage:
— Parental combinations (Round Yellow and Wrinkled Green) would appear in HIGHER frequency.
— Recombinant combinations (Round Green and Wrinkled Yellow) would appear in LOWER frequency (less than 25% each). [½ mark]

• Reason: When genes are linked on the same chromosome, they tend to be inherited together. Only occasional crossing over (recombination) between the two gene loci during meiosis produces the non-parental (recombinant) types. Therefore, the 1:1:1:1 ratio of the independent assortment test-cross is NOT obtained; instead, parental types predominate over recombinant types. [½ mark]
Q19Short Answer3 marks

In Antirrhinum (snapdragon), flower colour shows incomplete dominance. When a pure-breeding red-flowered plant is crossed with a pure-breeding white-flowered plant, the F₁ plants all bear pink flowers. (i) What are the genotype and phenotype of the F₁ plants? (ii) Write the cross showing the genotypes of all plants when F₁ pink-flowered plants are selfed (F₁ × F₁). Show this cross using a Punnett square. (iii) State the phenotypic ratio of the F₂ generation.

Show answer
Part (i): Genotype and phenotype of F₁ plants ½ + ½ = 1 mark

Parental Cross:
- Pure-breeding red: R¹R¹
- Pure-breeding white: R²R²

F₁ genotype: R¹R²
F₁ phenotype: Pink (intermediate phenotype — neither allele is completely dominant; this is incomplete dominance)

---

Part (ii): Punnett square for F₁ × F₁ (selfing of pink) — 1 mark

Parents (F₁): R¹R² × R¹R²

Gametes of each F₁ parent: R¹ and R²

| | | |
|---|---|---|
| | R¹R¹ | R¹R² |
| | R¹R² | R²R² |

F₂ genotypes obtained:
- R¹R¹ (pure red)
- R¹R² (pink) — appears twice
- R²R² (pure white)

Genotypic ratio: 1 R¹R¹ : 2 R¹R² : 1 R²R²

---

Part (iii): Phenotypic ratio of F₂ generation — 1 mark

| Genotype | Phenotype |
|---|---|
| R¹R¹ | Red |
| R¹R² | Pink |
| R²R² | White |

Phenotypic ratio: 1 Red : 2 Pink : 1 White

*(Note: In incomplete dominance, the phenotypic ratio equals the genotypic ratio — 1:2:1 — unlike Mendelian dominance where F₂ gives 3:1.)*
Q20Short Answer3 marks

A genetics student is analysing data from two separate crosses in Drosophila melanogaster:

Cross 1: A homozygous wild-type female (grey body, normal wings) was crossed with a male having black body and vestigial wings. The F1 females were then test-crossed with black-bodied, vestigial-winged males. Expected result if genes are unlinked: 1:1:1:1. Observed result: predominantly parental combinations with only a small percentage of recombinant classes.

Cross 2: A human carrier female (X^H X^h) married a normal male (X^H Y). Their son was found to be haemophilic.

(i) Explain why Cross 1 does NOT give a 1:1:1:1 ratio, naming the phenomenon responsible and the scientist associated with it. (1 mark)
(ii) Using the Cross 2 data, draw the correct Punnett square and identify the genotype(s) of daughters who would be carriers. (1½ marks)
(iii) In Cross 1, if the recombination frequency between the two loci is found to be 17%, what does this value indicate about the relative positions of the two genes on the chromosome? (½ mark)

Show answer
Answer (3 marks total)

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Part (i) — 1 mark
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
• The two genes (body colour and wing type) are located on the SAME chromosome — they are LINKED genes.
• Phenomenon: LINKAGE (incomplete linkage / crossing over)
• Scientist: T.H. Morgan (worked with Drosophila; coined the term linkage).
• Because the genes do not assort independently (violates Mendel's Law of Independent Assortment), the parental combinations are more frequent than recombinant combinations → ratio deviates from expected 1:1:1:1.
[Award 1 mark: linkage named + Morgan credited; partial credit ½ mark if only one of the two is stated]

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Part (ii) — 1½ marks
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Notation key:
X^H = dominant allele (normal clotting)
X^h = recessive allele (haemophilia)

Parents:
Mother (carrier female): X^H X^h
Father (normal male): X^H Y

Gametes:
Mother produces: X^H and X^h
Father produces: X^H and Y

Punnett Square:

┌──────────────┬──────────────┬──────────────┐
│ │ X^H │ Y │
├──────────────┼──────────────┼──────────────┤
│ X^H │ X^H X^H │ X^H Y │
│ │(Normal ♀) │(Normal ♂) │
├──────────────┼──────────────┼──────────────┤
│ X^h │ X^H X^h │ X^h Y │
│ │(Carrier ♀) │(Haemo. ♂) │
└──────────────┴──────────────┴──────────────┘

• The haemophilic son has genotype X^h Y — he received X^h from his carrier mother and Y from his father. ✓

• Daughters who are CARRIERS: genotype X^H X^h
→ 50% of daughters (1 out of 2 daughters) will be carriers.

[Award ½ mark: correct Punnett square layout with parental gametes on top and side; ½ mark: correct genotypes in all 4 cells; ½ mark: correctly identifying X^H X^h as carrier daughter]

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Part (iii) — ½ mark
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
• Recombination frequency (RF) = 17% means the two genes are 17 map units (centimorgans, cM) apart on the same chromosome.
• RF < 50% confirms the genes ARE linked (not on separate chromosomes).
• The closer to 0%, the tighter the linkage; the closer to 50%, the more independent the assortment appears.
• Conclusion: body colour and wing-type loci are 17 cM apart on chromosome 2 of Drosophila — showing INCOMPLETE LINKAGE with crossing over occurring between them.
[Award ½ mark: states 17 map units / 17 cM distance OR states genes are linked and RF indicates relative distance on same chromosome]

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
MARKING SUMMARY
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
(i) Linkage + Morgan credited = 1 mark
(ii) Correct Punnett square + carrier identified = 1½ marks
(iii) RF = map distance + linkage interpretation = ½ mark
TOTAL = 3 marks
Q21Short Answer3 marks

A woman with normal vision, whose father was colour blind, marries a man with normal vision. (i) Give the genotypes of the woman and her father. (ii) Work out a cross to show the possible genotypes and phenotypes of their sons and daughters. (iii) A son from this marriage marries a colour-blind woman. What is the probability that their daughter will be colour blind? Give reason.

Show answer
Part (i): Genotypes — 1 mark

Colour blindness is an X-linked recessive trait. Let X^C = normal vision allele (dominant); X^c = colour-blind allele (recessive).

• Father (colour blind): X^c Y
• The woman received X^c from her colour-blind father (a daughter always gets one X from her father) and X^C from her normal-visioned mother.
→ Woman's genotype: X^C X^c (carrier female — normal vision)

---

Part (ii): Cross — 1 mark

Parents:
| | Mother (carrier) X^C X^c | | Father (normal) X^C Y |
|---|---|---|---|

Gametes of mother: X^C , X^c
Gametes of father: X^C , Y

Punnett Square:

| | X^C (father) | Y (father) |
|---|---|---|
| X^C (mother) | X^C X^C | X^C Y |
| X^c (mother) | X^C X^c | X^c Y |

Results:

| Genotype | Phenotype |
|---|---|
| X^C X^C | Normal daughter (homozygous) |
| X^C X^c | Carrier daughter (normal vision) |
| X^C Y | Normal son |
| X^c Y | Colour-blind son |

• Daughters: 1/2 normal (X^C X^C) : 1/2 carrier (X^C X^c) — all daughters have normal vision
• Sons: 1/2 normal (X^C Y) : 1/2 colour blind (X^c Y)

---

Part (iii): Probability of colour-blind daughter — 1 mark

A colour-blind son from above cross has genotype: X^c Y
He marries a colour-blind woman: genotype X^c X^c

Cross: X^c Y × X^c X^c

| | X^c (mother) | X^c (mother) |
|---|---|---|
| X^c (father) | X^c X^c | X^c X^c |
| Y (father) | X^c Y | X^c Y |

All daughters receive X^c from father and X^c from mother → genotype X^c X^c (colour blind).

Probability that their daughter is colour blind = 1 (100%)

Reason: The father (X^c Y) can only pass X^c to all daughters; the colour-blind mother (X^c X^c) also passes only X^c. Therefore every daughter must be homozygous X^c X^c and will express colour blindness.
Q22Long Answer5 marks

A woman with normal vision, whose father was colour blind, marries a man with normal vision. With the help of a cross, determine the genotypes of the parents and work out the probability of their children being colour blind. Also, (i) state the pattern of inheritance of colour blindness in humans, and (ii) explain why sons are more frequently affected than daughters in X-linked recessive disorders.

Show answer
Answer:

Determining the genotypes of the parents: (1 mark)

Colour blindness is caused by a recessive allele (X^b) located on the X chromosome.
- The woman has normal vision but her father was colour blind.
- Her father's genotype = X^b Y (colour blind male).
- She must have received X^b from her father and X^B from her normal mother.
- Therefore, the woman is a carrier: genotype = X^B X^b
- The man has normal vision: genotype = X^B Y

Cross: (2 marks)

```
Parents: Carrier Female × Normal Male
Genotype: X^B X^b × X^B Y

Gametes: X^B X^b X^B Y
```

Punnett Square:

| | X^B (from father) | Y (from father) |
|---------------|-------------------|------------------|
| X^B (from mother) | X^B X^B | X^B Y |
| X^b (from mother) | X^B X^b | X^b Y |

Offspring genotypes and phenotypes:

| Genotype | Phenotype |
|----------|-----------|
| X^B X^B | Normal vision female (homozygous dominant) |
| X^B X^b | Carrier female (normal vision) |
| X^B Y | Normal vision male |
| X^b Y | Colour blind male |

Probability of colour blind children:
- Probability that a child is colour blind = 1 out of 4 (25%)
- Probability that a son is colour blind = 1 out of 2 sons (50%)
- Probability that a daughter is colour blind = 0 (0%) — daughters cannot be colour blind in this cross

(i) Pattern of inheritance of colour blindness: (1 mark)

Colour blindness follows X-linked recessive inheritance.
- The gene for colour vision is located on the X chromosome (sex-linked).
- The recessive allele (X^b) causes colour blindness only when present in hemizygous condition in males (X^b Y) or homozygous condition in females (X^b X^b).
- It shows criss-cross inheritance — an affected father passes the recessive allele to all his daughters (who become carriers), NOT to his sons.

(ii) Why sons are more frequently affected than daughters: (1 mark)

- Males are hemizygous for the X chromosome (genotype: XY) — they have only ONE X chromosome.
- A single copy of the recessive allele X^b on the X chromosome is sufficient to cause colour blindness in males, because there is NO second X chromosome to carry the dominant allele X^B to mask its effect.
- Females have TWO X chromosomes (XX); to be colour blind, a female must inherit the recessive allele X^b from BOTH parents (genotype X^b X^b), which is a far less frequent event.
- Therefore, sons are affected much more frequently than daughters in all X-linked recessive disorders.

(1×5 = 5 marks)
Q23Long Answer5 marks

A pure breeding tall pea plant (Pisum sativum) is crossed with a pure breeding dwarf pea plant. Separately, a pure breeding red-flowered Mirabilis jalapa (four o'clock plant) is crossed with a pure breeding white-flowered Mirabilis jalapa. (a) Work out the F₁ and F₂ generations for BOTH crosses using appropriate Punnett squares. (b) State the genotypic ratio and phenotypic ratio obtained in each F₂ generation. (c) Comment on the difference in the pattern of inheritance observed in the two crosses, naming the phenomenon responsible in each case.

Diagram for question 23: Principles of Inheritance and Variation
Show answer
CROSS 1 — Pea Plant (Tall × Dwarf): Dominance

Parents (P):
Tall (TT) × Dwarf (tt)

Gametes:
T (all) × t (all)

F₁ Generation:
All offspring = Tt (Tall)
→ Phenotype: 100% Tall
→ The dwarf trait is not expressed in F₁ — it is MASKED by the dominant allele T.

F₁ × F₁ Cross (Tt × Tt):

Gametes of F₁ Tall (Tt): T and t

PUNNETT SQUARE — F₂ (Cross 1):

T t
┌──────────┬──────────┐
T │ TT │ Tt │
│ Tall │ Tall │
├──────────┼──────────┤
t │ Tt │ tt │
│ Tall │ Dwarf │
└──────────┴──────────┘

F₂ Genotypic Ratio: 1 TT : 2 Tt : 1 tt
F₂ Phenotypic Ratio: 3 Tall : 1 Dwarf

────────────────────────────────────────

CROSS 2 — Mirabilis jalapa (Red × White): Incomplete Dominance

Parents (P):
Red-flowered (R¹R¹) × White-flowered (R²R²)
[Note: Neither allele is completely dominant over the other.]

Gametes:
R¹ (all) × R² (all)

F₁ Generation:
All offspring = R¹R² (PINK)
→ Phenotype: 100% Pink
→ Neither red nor white is expressed completely; the heterozygote shows an INTERMEDIATE (blended) phenotype — PINK.

F₁ × F₁ Cross (R¹R² × R¹R²):

Gametes of F₁ Pink (R¹R²): R¹ and R²

PUNNETT SQUARE — F₂ (Cross 2):

R¹ R²
┌───────────┬───────────┐
R¹ │ R¹R¹ │ R¹R² │
│ Red │ Pink │
├───────────┼───────────┤
R² │ R¹R² │ R²R² │
│ Pink │ White │
└───────────┴───────────┘

F₂ Genotypic Ratio: 1 R¹R¹ : 2 R¹R² : 1 R²R²
F₂ Phenotypic Ratio: 1 Red : 2 Pink : 1 White

────────────────────────────────────────

(c) COMMENT ON PATTERNS OF INHERITANCE:

• Cross 1 (Pea — Tall × Dwarf) shows the Law of Dominance as proposed by Gregor Mendel. One allele (T = Tall) is completely dominant over the other allele (t = Dwarf). In F₁, only the dominant phenotype appears. In F₂, the phenotypic ratio is 3:1, which is DIFFERENT from the genotypic ratio (1:2:1). The recessive phenotype reappears in F₂ unchanged — it was only masked, not lost.

• Cross 2 (Mirabilis jalapa — Red × White) shows INCOMPLETE DOMINANCE — neither allele completely dominates the other. The F₁ heterozygote (R¹R²) shows a phenotype INTERMEDIATE between the two parents (pink). In F₂, the phenotypic ratio (1:2:1) is IDENTICAL to the genotypic ratio (1:2:1) — this is a key distinguishing feature of incomplete dominance. The original parental phenotypes (red and white) reappear in F₂ along with the intermediate pink, proving the alleles remain discrete and do NOT blend permanently (particulate inheritance is maintained).

• KEY DISTINCTION: In dominance (Cross 1), phenotypic ratio ≠ genotypic ratio in F₂ (3:1 vs 1:2:1). In incomplete dominance (Cross 2), phenotypic ratio = genotypic ratio in F₂ (both 1:2:1).
Q24Long Answer5 marks

A genetics counsellor is studying the inheritance of a rare autosomal recessive disorder called 'Maple Syrup Urine Disease' (MSUD) in a family. The pedigree chart of the family is given below:

Generation I: Carrier female (I-1) × Normal male (I-2)
Generation II: II-1 (affected male), II-2 (normal female — married into family), II-3 (carrier female), II-4 (normal male — married into family)
Generation III: III-1 (normal female), III-2 (affected male), III-3 (normal male)

[Note: Affected individuals are shown as filled symbols; carriers are half-filled; normal individuals are unfilled. Square = male, Circle = female]

The counsellor informs the family that MSUD is caused by a mutation in a gene encoding a metabolic enzyme, and that the disorder follows Mendelian inheritance.

(a) Identify the type of inheritance shown in this pedigree. Give TWO reasons to justify your answer. (2 marks)
(b) Write the genotypes of individuals I-1, I-2, and II-3. Use 'M' for the dominant allele and 'm' for the recessive allele. (1 mark)
(c) The couple II-3 and II-4 (II-4 is a normal, non-carrier male) plan to have a fourth child. Using a Punnett square, show the possible genotypes of their offspring and state the probability that the fourth child will be an affected male. (2 marks)

Diagram for question 24: Principles of Inheritance and Variation
Show answer
MARKING SCHEME — Principles of Inheritance and Variation (5 marks)

─────────────────────────────────────────
(a) Type of inheritance and justification: (1 × 2 = 2 marks)
─────────────────────────────────────────

Type: AUTOSOMAL RECESSIVE inheritance. (stated as part of justification — no separate mark)

Reason 1 (1 mark):
Both parents of Generation I (I-1 and I-2) are unaffected (normal/carrier), yet their son II-1 is affected. This means the disorder skips a generation and unaffected parents can produce affected offspring — a hallmark of RECESSIVE inheritance.

Reason 2 (1 mark):
The disorder affects both males (II-1, III-2) and females could also be affected (equal probability for both sexes), and the causative gene is present in Generation I parents who are phenotypically normal. This rules out X-linked inheritance (since II-1 would need to inherit the allele from his father I-2, who is normal and unaffected, and X-linked recessive would require I-2 to be affected). Hence the gene is located on an AUTOSOME.

[Acceptable alternative for Reason 2: The affected son II-1 gets the recessive allele from his father I-2 who is phenotypically normal — if X-linked, I-2 would himself be affected since males have only one X chromosome. Therefore, the gene must be autosomal.]

─────────────────────────────────────────
(b) Genotypes of I-1, I-2, and II-3: (1 mark for all three correct)
─────────────────────────────────────────

• I-1 (Carrier female): Mm
• I-2 (Normal male): Mm [must be carrier since affected son II-1 = mm; I-2 is phenotypically normal but must contribute 'm' allele]
• II-3 (Carrier female): Mm

(Award 1 mark if all three genotypes are correctly stated; deduct if any one is wrong)

─────────────────────────────────────────
(c) Punnett square and probability: (1 + 1 = 2 marks)
─────────────────────────────────────────

Parents:
• II-3 (carrier female) = Mm
• II-4 (normal, non-carrier male) = MM

Cross: Mm × MM

Gametes of II-3: M and m
Gametes of II-4: M and M

PUNNETT SQUARE:

┌──────────────┬──────────────┬──────────────┐
│ │ M (II-4) │ M (II-4) │
├──────────────┼──────────────┼──────────────┤
│ M (II-3) │ MM │ MM │
├──────────────┼──────────────┼──────────────┤
│ m (II-3) │ Mm │ Mm │
└──────────────┴──────────────┴──────────────┘

Offspring genotypes:
• MM (normal, non-carrier) — 2 out of 4 = 50%
• Mm (carrier, phenotypically normal) — 2 out of 4 = 50%
• mm (affected) — 0 out of 4 = 0%

Genotypic ratio: 2 MM : 2 Mm (i.e., 1 MM : 1 Mm)
Phenotypic ratio: All offspring phenotypically NORMAL (no affected child possible)

Probability that the fourth child will be an AFFECTED MALE (mm AND male):
= Probability of mm genotype × Probability of being male
= 0 × 1/2
= 0 (ZERO probability)

The fourth child CANNOT be affected because II-4 does not carry the recessive allele 'm'; therefore no offspring can be homozygous recessive (mm). (1 mark for correct Punnett square with gametes labelled; 1 mark for correct probability statement with reasoning)

─────────────────────────────────────────
SUMMARY OF VALUE POINTS:
─────────────────────────────────────────
(a) Reason 1 — unaffected parents produce affected child (skip generation) .............. 1 mark
(a) Reason 2 — not X-linked because affected males receive allele from normal father ... 1 mark
(b) All three genotypes correct: I-1 = Mm, I-2 = Mm, II-3 = Mm ......................... 1 mark
(c) Correct Punnett square with parental gametes shown on rows/columns .................. 1 mark
(c) Probability = 0 with correct justification (II-4 is MM, so mm impossible) .......... 1 mark
─────────────────────────────────────────
TOTAL: 5 marks
Q25Long Answer5 marks

A geneticist studied a three-generation family in which some members showed a rare autosomal recessive condition called phenylketonuria (PKU). The pedigree is described below:

Generation I: Individual 1 (unaffected male) × Individual 2 (unaffected female)
Generation II: Individual 3 (unaffected male), Individual 4 (affected female), Individual 5 (unaffected female), Individual 6 (unaffected male)
Individual 5 (Generation II) × Individual 6 (Generation II) are married
Generation III: Individual 7 (affected male), Individual 8 (unaffected female), Individual 9 (unaffected male)

Using the information above, answer the following:
(a) Is PKU autosomal or X-linked? Give one reason from the pedigree to justify your answer. (1 mark)
(b) Give the genotypes of Individual 1, Individual 2, Individual 5, and Individual 6. Use 'P' for the dominant allele and 'p' for the recessive allele. (2 marks)
(c) What is the probability that Individual 8 is a carrier of PKU? Show the cross between Individuals 5 and 6 using a Punnett square to support your answer. (2 marks)

Show answer
(a) PKU is AUTOSOMAL. (1 mark)
Justification: Individual 4 (Generation II) is an affected female. If PKU were X-linked recessive, she would need to inherit the recessive allele from her father (Individual 1) on his X chromosome, making Individual 1 affected. However, Individual 1 is unaffected. Therefore, PKU cannot be X-linked; it must be autosomal.

(b) Genotypes of key individuals: (1 × 2 = 2 marks)
• Individual 1: Pp
• Individual 2: Pp
• Individual 5: Pp
• Individual 6: Pp

Reasoning: Individual 4 is affected (pp); both her parents (Individuals 1 and 2) are unaffected yet must each carry one 'p' allele, so both are Pp. Individual 7 (Generation III) is affected (pp); both his parents (Individuals 5 and 6) must each carry one 'p' allele. Individual 5 is an unaffected daughter of Pp × Pp parents who has an affected child, confirming she is Pp. Individual 6 is unaffected but has an affected child, confirming he is Pp.
[Award 1 mark for any two correct genotypes; 2 marks for all four correct genotypes]

(c) Probability that Individual 8 is a carrier + Punnett square: (1 + 1 = 2 marks)

Cross: Individual 5 (Pp) × Individual 6 (Pp)

Punnett Square: (1 mark)

P (from Ind. 6) p (from Ind. 6)
P (from Ind. 5) PP Pp
p (from Ind. 5) Pp pp

Offspring genotype ratio:
• PP : Pp : pp = 1 : 2 : 1
• Phenotype ratio: 3 unaffected : 1 affected

Probability that Individual 8 is a carrier: (1 mark)
Individual 8 is stated to be unaffected. Among unaffected offspring, the possible genotypes are PP (1) and Pp (2), giving a total of 3 unaffected individuals.
Out of these 3 unaffected individuals, 2 are carriers (Pp).
Therefore, the probability that Individual 8 (an unaffected individual) is a carrier = 2/3.

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Principles of Inheritance and Variation Class 12 Questions